1
$\begingroup$

Let $f: C \rightarrow I$ map each point of the middle third Cantor set $C$, expressed as a ternary number which contains only digits $0$ and $2$, to the set of real numbers in $I=[0,1]$ expressed in base $2$ according the the rule:

$0.a_1a_2a_3...\rightarrow 0.b_1b_2b_3...$

where $b_i = \frac{a_i}{2}$.

How do I prove this map is continuous?

$\endgroup$
  • $\begingroup$ Wikipedia claims, Cantor set is a metric space equipped with inherited topology, and you can easily show, it is a Lipschitz with a constant $\frac{1}{2}$. $\endgroup$ – dEmigOd Apr 24 '18 at 7:59
  • $\begingroup$ @dEmigOd Your claim is not correct, because $f(1)-f(0)=1$. In fact, the map sends hollow intervals of $C$ of length $3^{-k}$ to full intervals of length $2^{-k}$, giving for each of those an "expansion factor" of at least $\frac{2^{-k}}{3^{-k}}=1.5^k$. So, the map is not Lipschitz. Perhaps it might be Holder, I don't know. $\endgroup$ – user228113 Apr 24 '18 at 8:24
  • $\begingroup$ My previous comment is wrong (since this is not decimal expansions!). But I don't see how you take something like $0.0222..._3$ and $0.2_3$ and send it to something of length $2^{-1}$ and not 0? $\endgroup$ – dEmigOd Apr 24 '18 at 10:05
  • $\begingroup$ @G.Sassatelli It should be Holder-continuous with exponent $\log 2/\log 3$. $\endgroup$ – Erick Wong Apr 24 '18 at 17:20
1
$\begingroup$

Fix a ternary expansion $$x = \frac{a_1}{3} + \frac{a_2}{3^2} + \frac{a_3}{3^3} + \ldots,$$ where $a_n = 0, 2$ for all $n$. Further, fix $\varepsilon > 0$. Choose an $m \in \mathbb{N}$ such that $2^{-m} < \varepsilon$. Note that two binary expansions are within $2^{-m}$ if they agree up to their $m$th bit. That is, necessarily, $$\left|\left(\frac{b_1}{2} + \ldots + \frac{b_m}{2^m} + \frac{b_{m+1}}{2^{m+1}}+\ldots\right) - \left(\frac{b_1}{2} + \ldots + \frac{b_m}{2^m} + \frac{b'_{m+1}}{2^{m+1}}+\ldots\right)\right| < 2^{-m} < \varepsilon.$$ Consider another ternary expansion $$y = \frac{b_1}{3} + \frac{b_2}{3^2} + \frac{b_3}{3^3} + \ldots.$$ If $x \neq y$, then the ternary expansions must differ (note: this is not true in general, but true in our case since we are taking only ternary expansions with $0$ and $2$ trits). Let $k$ be the first digit that differs, that is, $b_k \neq a_k$, but $b_n = a_n$ for $n < k$. Then $|a_k - b_k| = 2$, and \begin{align*} |x - y| &\ge \frac{|a_k - b_k|}{3^k} - \frac{|a_2 - b_2|}{3^{k+1}} - \frac{|a_3 - b_3|}{3^{k+2}} - \ldots \\ &\ge \frac{2}{3^k} - \frac{2}{3^{k+1}} - \frac{2}{3^{k+2}} - \ldots = \frac{1}{3^k}. \end{align*} Therefore, if we set $\delta = 3^{-m}$, then $|y - x| < \delta$ implies that the ternary expansion agrees at least to $m$ trits. Thus, $f(y) - f(x)$ agree to $m$ bits, hence $|f(y) - f(x)| < 2^{-m} < \varepsilon$. This proves continuity.

$\endgroup$
0
$\begingroup$

For $x\in C$ and $\epsilon >0$, take $n\in \Bbb N $ with $2^{-n}<\epsilon$ and let $\delta=3^{-n}.$

If $x\ne y\in C$ let the $k$th digit-place be the least place (to the right of the decimal point) where their digits of $x,y$ in base-$3$ are unequal. Then $|x-y|\ge 2\cdot 3^{-k}-(0\cdot 3^{-k}+\sum_{j=1}^{\infty}2\cdot 3^{-k-j})=3^{-k}.$

So if $ y\in (-\delta+x,\delta+x)\cap C$ then $k\geq n,$ so $f(x),f(y)$ have the same first $n$ digits in base-$2$, so $|f(x)-f(y)|\leq 2^{-n}<\epsilon.$

$\endgroup$
  • $\begingroup$ The graph of $f$ has been called the Devil's Staircase because it is differentiable with derivative $0,$ except on $C$ (a set of measure $0$) yet $f$ is not constant. $\endgroup$ – DanielWainfleet Apr 24 '18 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.